The statements made herein merely provide information related to the present disclosure and may not constitute prior art, and may describe some embodiments illustrating the invention.
Measured nuclear magnetic resonance (NMR) data resulting from a multi-component sample can be denoted by G(t) which represents a multi-exponential decay, with time constants T2 and amplitudes ƒ(T2)G(t)=∫0∞Pτ(T2)e−t/T2ƒ(T2)dT2  (1)where the function Pτ(T2) is referred to as the polarization factor and depends on the pulse sequence of the NMR equipment used to probe and measure the sample.
In certain samples containing hydrocarbons and water, e.g., geological formations, the transverse relaxation time T2 is the characteristic de-phasing time for protons in hydrocarbons or water present in pores of a rock or in the bulk fluid. As aforementioned, the function Pτ(T2) of equation (1) depends on the pulse sequence of the NMR equipment used to probe and measure the sample. For example,
                                          P            τ                    ⁡                      (                          T              2                        )                          =                  {                                                    1                                                              CPMG                  ⁢                                                                          ⁢                  pulse                  ⁢                                                                          ⁢                  sequence                  ⁢                                                                          ⁢                  with                  ⁢                                                                          ⁢                  full                  ⁢                                                                          ⁢                  polarization                                                                                                      1                  -                                      2                    ⁢                                          ⅇ                                                                        -                          τ                                                /                                                  T                          2                                                                                                                                                                                      inversion                    ⁢                                                                                  ⁢                    recovery                                    -                                      CPMG                    ⁢                                                                                  ⁢                    pulse                    ⁢                                                                                  ⁢                    sequence                                                                                                                        1                  -                                      ⅇ                                                                  -                        τ                                            /                                              T                        2                                                                                                                                                              saturation                    ⁢                                                                                  ⁢                    recovery                                    -                                      CPMG                    ⁢                                                                                  ⁢                    pulse                    ⁢                                                                                  ⁢                    sequence                                                                                                          (        2        )            where τ is a function of pre-polarization time and longitudinal relaxation, and CPMG refers to the well-known Carr-Purcell-Meiboom-Gill sequence. In downhole applications, the function Pτ(T2) is a complex function of tool geometry (such as length of magnet and design of RF antenna), operational constraints (such as cable speed) as well as the pulse sequence.
In equation (1) the T2 distribution denoted by ƒ(T2) is estimated from indications of the measured data G(t). The conventional approach to estimating ƒ(T2) utilizes an inverse Laplace transform (ILT), See L. Venkataramanan et al., “Solving Freholm integrals of the first kind with tensor product structure in 2 and 2.5 dimensions”, IEEE Transactions on Signal Processing, 50:1017-1026, 2002. More particularly, it is assumed that the indications of data G(t) are measured with additive, white, Gaussian noice. Traditionally, assuming Pτ(T2) is known, an inversion algorithm is used to estimate the distribution of relaxation times ƒ(T2) in equation (1) from the measured data indications G(t). Next, linear functionals of the estimated ƒ(T2) can be used to estimate the petro-physical or fluid properties. For example, the area under the T2 distribution can be interpreted as the porosity of the rock. Often, based on lithology, a threshold T2 is chosen as the cut-off characteristic time separating fluid bound to the pore surface and fluid that is not bound to the pore surface and flow more easily. For example, in sandstones, the area under the T2 distribution corresponding to relaxation times smaller than 33 msec has been empirically related to bound fluid volume. The remaining area under the T2 distribution corresponds to free fluid volume. The mean of the distribution lnT2, is empirically related to either rock permeability and/or to hydrocarbon viscosity. The width of the distribution, σln(T2), provides a qualitative range of the distribution of pore sizes in the rock. Moments of relaxation time or diffusion are often related to rock permeability and/or hydrocarbon viscosity. Similar answer products can also be obtained from linear functionals computed from two-dimensional diffusion-relaxation data or T1-T2 relaxation data.
It is well known in the literature that estimation of ƒ(T2) is an ill-conditioned and non-linear problem. Small changes in the indications of measured data G(t) due to noise can result in widely different ƒ(T2). In theory, there are infinitely many ƒ(T2) that fit the data; i.e., there are non-unique solutions. In the literature, this problem is addressed by introducing a choosing the “smoothest” solution ƒ(T2) that fits the data. More particularly, a regularization functional is introduced, and the solution is estimated through minimization of a cost function Q with respect to the underlying distribution f,Q=∥G−Lƒ∥2+α∥ƒ∥2  (3)where G is a vector representing the measured data, L is the matrix of the discretized function Pτ(T2)e−t/T2, and f is the discretized version of the underlying density function ƒ(T2). The first term in the cost function is the least squares error between the data and the fit from the model in equation (1). The second term is referred to as the regularization and incorporates smoothness in the relaxation amplitudes into the problem formulation. The parameter α denotes the compromise between the fit to the data and an a priori expectation of the distribution. In equation (3), α is the weight given to the regularization and can be chosen by a number of different methods.
Previously incorporated U.S. Ser. No. 13/333,232 describes a method based on integral transforms that allows direct computation of a linear functional of ƒ(x) (where the variable x may be the transverse relaxation time (T2), the longitudinal relaxation time (T1), Diffusion (D), any of various two-dimensional data sets (such as D-T2 or D-T1 or T1-T2), multi-dimensional data sets (such as D-T2-T1, D-T2-T1-azimuth, D-T2-T1-different depth of investigations) without involving computation of ƒ(x). The central idea of this approach is to compute the integral transform of G(t) using a function k(t) such that the Laplace transform of k(t) denoted by K(T2) has the desired properties in the x domain (hereinafter for simplicity purposes described with respect to the T2 domain). Let the integral transform of the data indications G(t) be denoted by ℑ{G(t)}=A, such thatℑ{G(t)}=A=∫0∞k(t)G(t)dt  (4)From equations (1) and (4),A=∫0∞K(T2)Pτ(T2)ƒ(T2)dT2  (5)where the functions k(t) and K(T2) form a Laplace-transform pair, withK(T2)≡∫0∞k(t)e−t/T2dt  (6)
From the right hand side of equation (5), for a desired linear transformation in the T2 domain, the integral transform approach constructs a function k(t) in the time-domain, so that the scalar product of the function with the measured data provides A, the parameter of interest.
Using this approach, direct computation of linear functionals of the T2 distribution may be made without use of the ILT method. For example, as described in previously incorporated U.S. Ser. No. 13/333,232, the Mellin, Exponential Haar, and Fourier-Mellin transform of magnetization data can be used to estimate moments, tapered areas, and the porosity of the distribution directly from the measured data.
More particularly, consider measured CPMG data G(t), its corresponding and unknown distribution ƒ(T2), and a tapered transition K(T2) for computing the desired tapered area of the T2 distribution. The function k(t) corresponding to this tapered function K(T2) may be shown. This function k(t) can be found using multiple methods: (1) using either an analytical form for k(t) (examples of this is shown in Table A), (2) using a numerical approach, (3) using a method of successive approximations, and/or (4) using convolution analysis. All these methods are described below. Once k(t) is estimated, a scalar product of k(t) with G(t) directly provides the tapered area A. Since this approach does not involve solving for ƒ(T2) and then estimating ∫0∞K(T2)ƒ(T2)dT2, it is more straight-forward and not susceptible to the subjectivity of traditional algorithms that involve inversion of an ill-conditioned and non-linear problem.
As discussed below, given a desired K(T2), the function k(t) can be computed in four ways: often, it can be computed analytically or numerically. It can also be computed using the method of successive approximations to K(T2). An alternate method of computing it involves taking advantage of the convolution-multiplication equivalence between the time and T2 domain. All four methods are described below.
For a desired K(T2), when the function k(t) exists analytically or can be computed numerically, the parameter A is obtained from equation (5). However, the function k(t) may not exist ∀K(T2). When it exists, it may also have infinite energy which can be related to infinite uncertainty in the estimated parameter A leading to instability in computing the parameter. Thus, the integral transform approach can provide insight into what linear functionals of the T2 distribution can be directly applied to the data G(t) and are stable in the context of providing low uncertainty in A.
The uncertainty in A can be quantified as a function of the signal-to-noise ratio (SNR) in the measured data. Let σε denote the standard deviation of the additive white Gaussian noise in the data. Equation (5) can be computed in the discrete-time domain asA=tEΣn=0Nk(ntE)G(ntE)  (6a)where tE denotes the sampling time or echo spacing. Therefore,σA2=σε2tE[Σn=0Nk2(ntE)tE]  (6b)
Equation 6b shows that when the function k(t) is square integrable, i.e, k(t) has finite energy E, where E=∫0∞k2(t)dt, then the uncertainty in A is always finite and directly related to the uncertainty in the measurement.
In the sub-sections below, tables are described of integral transforms developed for different polarization factors PτT2 encountered in oilfield NMR applications.
A. Tapered Areas from Fully Polarized Data
Consider NMR data that have been fully polarized, with Pτ(T2)=1∀T2. In this sub-section, a few integral transforms are described where K(T2) corresponds to tapered and sharp Heaviside functions.
Let Tc denote the T2 relaxation time at which the desired cut-off of the tapered Heaviside function is 0.5. The parameter Tc is user-specified and may come from laboratory study of rock and fluid properties or may correspond to a value of T2 expected to separate two fluids in the T2 domain. For example, in sandstones and carbonates, the area under the T2 distribution corresponding to relaxation times smaller than 33 and 100 msec, respectively, has been empirically related to bound fluid volume. Thus, given a value of Tc, a tapered or sharp Heaviside function K(T2, Tc) is sought such that the tapered area can be computed in the time-domain using the corresponding function k(t, Tc). The integral transform for computing tapered and sharp transitions should satisfy the following properties:
1. The function k(t, Tc) should exist ∀t and K(T2, Tc) should exist ∀T2.
2. Based on the underlying petrophysics, it is desirable that K(T2, Tc) be monotonic between 0 and 1 (on the y-axis), withK(T2,Tc)|T2−0=0  (6c)limT2→∞K(T2,Tc)=1  (6d)K(T2,Tc)|T2−Tc=0.5  (6e)3. It should be possible to adjust the slope m in the log(T2) space of the transition region, with
                              m          ≡                      (                                          ⅆ                                  K                  ⁡                                      (                                                                  T                        2                                            ,                                              T                        c                                                              )                                                                                                ⅆ                  log                                ⁢                                                                  ⁢                                  T                  2                                                      )                          ⁢                  |                                    T              2                        =                          T              c                                                          (                  6          ⁢          f                )            
In most oilfield applications, the desired slope varies from m=0.4 for gradual tapered cut-offs to m=4 for sharp cut-offs.
Using analytical means a set of integral transforms are developed that satisfy these properties, and they are summarized in Table A. For ease of reference, suggested names for the transforms are provided based on the function k(t). The energy for some of the transforms is infinity, implying infinite uncertainty in the estimated area. This energy can be decreased by several methods. One such method involves multiplication of the function k(t) by an exponential decaying signal in the time domain. A second method involves restricting the integral transform to a finite time-period. Both methods decrease the energy considerably while also reducing the slope in the transition region. For e.g., as shown in Table A, the Haar transform (HT) (row 3) has infinite energy. On the other hand, the energy of an exponential Haar transform (EHT) (row 5) is finite.
From equation (7), a desired uncertainty in the estimated area σA can be translated to a desired and finite energy in the function. This finite energy can be achieved by suitable choice of parameters of the transform satisfying both the energy criteria as well as properties 1-3 described above. For example, when the desired energy for the EHT is
      2          π      ⁢                          ⁢              T        c              ,then me parameters C, α and β take values provided in the following table.
TABLE AIntegral transforms that give rise to tapered transitions in the log(T2) domainName ofK(T2, Tc)Parametersk(t, Tc)|EmTransform      2    π    ⁢            tan      ⁢                                  -      1        ⁢      (          αT      2        )    α  =      1          T      c            2    π    ⁢            sin      ⁡              (        αt        )              t    2      π    ⁢                  ⁢    T  0.32Sinc             T      2        α    ⁢      tanh    ⁡          (              α                  T          2                    )        α  =            T      c        0.52219              1      α        ⁢                  (                  -          1                )            n                  2      ⁢      n      ⁢                          ⁢      α        <    t    <          2      ⁢              (                  n          +          1                )            ⁢      α      ∞0.42Haar       α    2              α      2        +          1              T        2        2              α  =      1          T      c      αsin(αt)∞0.5Sine       C          (                        1                      T            2                          +        β            )        ⁢      tanh    ⁡          [              α        ⁡                  (                                    1                              T                2                                      +            β                    )                    ]            C    =          0.7213              T        c                  α    =                  (        1.572        )            ⁢              T        c                  β    =          0.4087              T        c            C(−1)ne−βt 2nα < t < 2(n + 1)α  2      π    ⁢                  ⁢          T      c      0.35Exponential Haar             α      2        +          β      2                  α      2        +                  (                  β          +                      1                          T              2                                      )            2            α    =                            4          ⁢          E          ⁢                                          ⁢          β                -                  β          2                          β    =          1                        T          c          2                ⁡                  (                                    4              ⁢              E                        -                          2                              T                c                                              )                                        α        2            +              β        2              α    ⁢      e                  -        β            ⁢                          ⁢      t        ⁢      sin    ⁡          (              α        ⁢                                  ⁢        t            )        2      π    ⁢                  ⁢          T      c      0.3Exponential Sine                     g        0            +                        ∑                      n            =            1                    ∞                ⁢                              a            n                    ⁢                                    g              n                        ⁡                          (              x              )                                            ,          x      =                        T          c                          T          2                                        g        n            ⁡              (        x        )              =                            x                                    2              ⁢              n                        -            2                          -                  x                      2            ⁢            n                                                (                      1            +                          x              2                                )                                      2            ⁢            n                    -          1                    ak computed recursivelySee Appendix A∞0.5 ≦ mn ≦ ∞ Variable SlopeSeries Expansion B. Transforms on Imperfectly Polarized Data
Consider imperfectly polarized data withPτ(T2)=1−e−τ/T2  (6g)where
      τ    =                  T        w            /      r        ,      r    =          〈                        T          1                          T          2                    〉      and Tw is the wait time. This polarization factor plays an important role in saturation-recovery-CPMG pulse sequence and in enhanced precision mode (EPM) used in downhole applications. We show below that the integral transform approach that was developed on fully polarized data with Pτ(T2)=1∀T2 can be applied to imperfectly polarized data as well. From eqns. (1) and (6g)
                              G          ⁡                      (            t            )                          =                                            ∫              0              ∞                        ⁢                                          ⅇ                                                      -                    t                                    /                                      T                    2                                                              ⁢                              f                ⁡                                  (                                      T                    2                                    )                                            ⁢                              ⅆ                                  T                  2                                                              -                                    ∫              0              ∞                        ⁢                                          ⅇ                                                      -                                          (                                              t                        +                        τ                                            )                                                        /                                      T                    2                                                              ⁢                              f                ⁡                                  (                                      T                    2                                    )                                            ⁢                              ⅆ                                  T                  2                                                                                        (                  6          ⁢          h                )            Let the fully polarized data be denoted by M(t), where
                              M          ⁡                      (            t            )                          =                              ∫            0            ∞                    ⁢                                    ⅇ                                                -                  t                                /                                  T                  2                                                      ⁢                          f              ⁡                              (                                  T                  2                                )                                      ⁢                                          ⅆ                                  T                  2                                            .                                                          (                  6          ⁢          i                )            Equation (6h) can then be cast as followsG(t)=M(t)−M(t+τ)  (6j)For a finite time t we have
                                                        lim                              N                ->                ∞                                      ⁢                          M              ⁡                              (                                  t                  +                                      N                    ⁢                                                                                  ⁢                    τ                                                  )                                              =          0.                ⁢                                  ⁢        Therefore                            (                  6          ⁢          l                )                                                      ∑                          n              =              0                                      N              -              1                                ⁢                      G            ⁡                          (                              t                +                                  n                  ⁢                                                                          ⁢                  τ                                            )                                      =                              M            ⁡                          (              t              )                                -                      M            ⁡                          (                              t                -                                  N                  ⁢                                                                          ⁢                  τ                                            )                                                          (                  6          ⁢          m                )            For a finite time t and in the limit N→∞, we get
                              M          ⁡                      (            t            )                          =                              ∑                          n              =              0                        ∞                    ⁢                                    G              ⁡                              (                                  t                  +                                      n                    ⁢                                                                                  ⁢                    τ                                                  )                                      .                                              (                  6          ⁢          n                )            
Therefore, if τ is either known or estimated, the fully polarized data M(t) can be reconstructed using equation (6n) from the measured data G(t). Integral transforms can be applied to M(t) to directly estimate linear functionals of ƒ(T2).
Effect of Noise
In practice we have a limited number of noisy samples of the form{tilde over (G)}(itE)=G(itE)+n(itE), i=1, . . . , N.  (6o)Using equation (6n) will result on a modified noise
                                          n            s                    ⁡                      (                          it              E                        )                          =                              ∑                          j              =              0                        N                    ⁢                      n            ⁡                          (                                                it                  E                                +                jτ                            )                                                          (                  6          ⁢          p                )            with varianceσM2=Nσε2.  (6q)For noisy data, it may be desirable to perform a denoising procedure before applying equation (6n).C. Transforms on Data from Logging Tools
The function k(t) can also be found using a combination of numerical and analytical methods as follows. This method is illustrated with an example in logging tools where the polarization factor Pτ(T2) tends to be more complex and depends on a number of parameters including hardware design such as length of permanent magnet, cable speed, etc. For simplicity, we have ignored the subscript τ in the polarization term in this sub-section. In logging applications, we have found that over a range of factors, P(T2) is well represented by
                              P          ⁡                      (                          T              2                        )                          =                  1                                    ∑                              k                =                0                            ∞                        ⁢                                          a                k                            ⁢                              ⅇ                                                      -                    bk                                    /                                      T                    2                                                                                                          (                  6          ⁢          r                )            Equation (6r) is a good fitting function ƒ or a wide range of parameters. For example, the imperfectly polarized data in the last sub-section is obtained from equation (6r) with b=τ and ak=1∀k. Similarly, at a range of cable speeds, the fit from equation (6r) matches P(T2) reasonably well. The polarization factor in logging tools is a complex function of tool geometry, operational constraints such as logging speed and and pulse sequence. In a number of circumstances, at logging speeds varying from 800-2000 ft/hour, the fit (solid line) from equation (6r) fits the polarization factor very well.
The fully polarized data M(t) can be reconstructed from G(t). From eqns. (6i) and (6n), we get
                                                                        M                ⁡                                  (                  t                  )                                            =                            ⁢                                                ∫                  0                  ∞                                ⁢                                                      ⅇ                                                                  -                        t                                            /                                              T                        2                                                                              ⁢                                      f                    ⁡                                          (                                              T                        2                                            )                                                        ⁢                                                            P                      ⁡                                              (                                                  T                          2                                                )                                                                                    P                      ⁡                                              (                                                  T                          2                                                )                                                                              ⁢                                      ⅆ                                          T                      2                                                                                                                                              =                            ⁢                                                ∫                  0                  ∞                                ⁢                                                      ⅇ                                                                  -                        t                                            /                                              T                        2                                                                              ⁢                                      f                    ⁡                                          (                                              T                        2                                            )                                                        ⁢                                                            P                      ⁡                                              (                                                  T                          2                                                )                                                              [                                                                  ∑                        k                                            ⁢                                                                        a                          k                                                ⁢                                                  ⅇ                                                                                    -                              bk                                                        /                                                          T                              2                                                                                                                                            ]                                    ⁢                                      ⅆ                                          T                      2                                                                                                                                              =                            ⁢                                                ∑                  k                                ⁢                                                      a                    k                                    ⁢                                                            ∫                      0                      ∞                                        ⁢                                                                  ⅇ                                                                              -                                                          (                                                              t                                +                                bk                                                            )                                                                                /                                                      T                            2                                                                                              ⁢                                              P                        ⁡                                                  (                                                      T                            2                                                    )                                                                    ⁢                                              f                        ⁡                                                  (                                                      T                            2                                                    )                                                                    ⁢                                              ⅆ                                                  T                          2                                                                                                                                                                                            =                            ⁢                                                ∑                  k                                ⁢                                                      a                    k                                    ⁢                                                            G                      ⁡                                              (                                                  t                          +                          bk                                                )                                                              .                                                                                                          (                  6          ⁢          s                )            
Integral transforms can be applied to M(t) to directly estimate linear functionals of ƒ(T2).
Method 2: The function k(t) can also be found numerically as follows. For example, it is possible that either Pτ(T2) or K(T2) is not well approximated by a closed form expression or an analytical k(t) does not exist for a specified K(T2). In this case, k(t) can be computed numerically as follows. For example, consider the case where the data are fully polarized with Pτ(T2)=1∀T2. The desired K(T2, Tc) can be shown as a trace. A numerical least squares approximation to k(t, T) can be obtained using singular value decomposition (SVD), with{tilde over (k)}(t)≈VnΣn−1UnTK(T2)  (6t)
Here matrices U, Σ and V are obtained by SVD of function e−t/T2 and n refers to the number of significant singular values.
In another embodiment of the method, the function k(t) can be found such that its Laplace transform minimizes the error with respect to the desired K(T2, Tc) and has a minimal energy,mink(t)∥∫0∞k(t)et/T2dt−K(T2,Tc)∥2 such that∥k∥2<E 
Method 3: The function k(t) can also be found using the equivalence of the convolution-multiplication operation between the time and T2 domain. This is further described below. We show that the product of two functions in the T2 domain corresponds to convolution in the time-domain. This property implies that the integral transforms described in this memo can also be combined in the time domain to estimate other parameters. For example, the moments of a specified region of the T2 distribution can be computed by using a function computed as the convolution of the Mellin operator and the Exponential Haar transform. Consider two different integral transforms of the measured data, where function k(t) in equation (5) is represented by k1(t) and k2(t) respectively,
                              A          1                =                                            ∫              0              ∞                        ⁢                                                            k                  1                                ⁡                                  (                  t                  )                                            ⁢                              G                ⁡                                  (                  t                  )                                            ⁢                              ⅆ                t                                              =                                    ∫              0              ∞                        ⁢                                                            K                  1                                ⁡                                  (                                      T                    2                                    )                                            ⁢                                                P                  τ                                ⁡                                  (                                      T                    2                                    )                                            ⁢                              f                ⁡                                  (                                      T                    2                                    )                                            ⁢                              ⅆ                                  T                  2                                                                                        (                  6          ⁢          u                )                                          A          2                =                                            ∫              0              ∞                        ⁢                                                            k                  2                                ⁡                                  (                  t                  )                                            ⁢                              G                ⁡                                  (                  t                  )                                            ⁢                              ⅆ                t                                              =                                    ∫              0              ∞                        ⁢                                                            K                  2                                ⁡                                  (                                      T                    2                                    )                                            ⁢                                                P                  τ                                ⁡                                  (                                      T                    2                                    )                                            ⁢                              f                ⁡                                  (                                      T                    2                                    )                                            ⁢                                                ⅆ                                      T                    2                                                  .                                                                        (                  6          ⁢          v                )            
Here, the functions K1(T2) and K2(T2) correspond to different linear functionals. Our interest is in evaluation of A3, where
                                          A            3                    =                                    ∫              0              ∞                        ⁢                                                            K                  3                                ⁡                                  (                                      T                    2                                    )                                            ⁢                                                P                  τ                                ⁡                                  (                                      T                    2                                    )                                            ⁢                              f                ⁡                                  (                                      T                    2                                    )                                            ⁢                              ⅆ                                  T                  2                                                                    ,                                  ⁢        and                            (                  6          ⁢          w                )                                                      K            3                    ⁡                      (                          T              2                        )                          ≡                                            K              1                        ⁡                          (                              T                2                            )                                ⁢                                    K              2                        ⁡                          (                              T                2                            )                                                          (                  6          ⁢          x                )                                =                              ∫            0            ∞                    ⁢                                                    k                1                            ⁡                              (                τ                )                                      ⁢                          ⅇ                                                -                  τ                                                  T                  2                                                      ⁢                          ⅆ              τ                        ⁢                                          ∫                0                ∞                            ⁢                                                                    k                    2                                    ⁡                                      (                                          t                      2                                        )                                                  ⁢                                  ⅇ                                                            -                                              t                        2                                                                                    T                      2                                                                      ⁢                                  ⅆ                                      t                    2                                                                                                          (                  6          ⁢          y                )            From equation (6)
                                                        ∫              0              ∞                        ⁢                                                            k                  3                                ⁡                                  (                  t                  )                                            ⁢                              ⅇ                                                      -                    t                                    /                                      T                    2                                                              ⁢                              ⅆ                t                                              =                                    ∫              0              ∞                        ⁢                                          ∫                0                ∞                            ⁢                                                                    k                    1                                    ⁡                                      (                    τ                    )                                                  ⁢                                                      k                    2                                    ⁡                                      (                                          t                      2                                        )                                                  ⁢                                  ⅇ                                                            -                                              (                                                  τ                          +                                                      t                            2                                                                          )                                                                                    T                      2                                                                      ⁢                                  ⅆ                  τ                                ⁢                                  ⅆ                                      t                    2                                                                                      ⁢                                  ⁢                                            Let              ⁢                                                          ⁢              τ                        +                          t              2                                =                      t            .                                                  ⁢            Thus                                              (                  6          ⁢          z                )                                                      ∫            0            ∞                    ⁢                                                    k                3                            ⁡                              (                t                )                                      ⁢                          ⅇ                                                -                  t                                /                                  T                  2                                                      ⁢                          ⅆ              t                                      =                              ∫            0            ∞                    ⁢                                    [                                                ∫                  0                  t                                ⁢                                                                            k                      1                                        ⁡                                          (                      τ                      )                                                        ⁢                                                            k                      2                                        ⁡                                          (                                              t                        -                        τ                                            )                                                        ⁢                                      ⅆ                    τ                                                              ]                        ⁢                          ⅇ                                                -                  t                                /                                  T                  2                                                      ⁢                          ⅆ              t                                                          (                  6          ⁢          aa                )            Thus, the parameter A3 can be computed as,
                              A          3                =                              ∫            0            ∞                    ⁢                                                    k                3                            ⁡                              (                t                )                                      ⁢                          G              ⁡                              (                t                )                                      ⁢                          ⅆ              t                                                          (                  6          ⁢          bb                )            where k3(t) is obtained as a convolution of k1(t) and k2(t),
                                          k            3                    ⁡                      (            t            )                          =                              ∫            0            t                    ⁢                                                    k                1                            ⁡                              (                τ                )                                      ⁢                                          k                2                            ⁡                              (                                  t                  -                  τ                                )                                      ⁢                                          ⅆ                τ                            .                                                          (                  6          ⁢          cc                )            
Hence, the product of two functions in the T2 domain corresponds to convolution in the time-domain. This property implies that the integral transforms described in this manuscript can also be combined to estimate other parameters. For example, this property implies that the moments of a specified region of the T2 distribution can be computed by integral transforms of the measured data, using a function obtained as a convolution of the Mellin operator and the Exponential Haar transform.
Method 4: We describe below a method for computing k(t) using method of successive approximations. We illustrate this with an example. Consider K(T2) is an arbitrarily sharp transition in the T2 domain. Let
  x  =                    T        c                    T        2              .  Let the Heaviside function H(x) be defined as follows.
                              H          ⁡                      (            x            )                          =                              1            ⁢                                                  ⁢            for            ⁢                                                  ⁢            x                    <          1                                    (                  6          ⁢          ee                )                                                          ⁢                  =                                    0.5              ⁢                                                          ⁢              for              ⁢                                                          ⁢              x                        =            1                                              (                  6          ⁢          ff                )                                                          ⁢                  =                      0            ⁢                                                  ⁢                          otherwise              .                                                          (                  6          ⁢          gg                )            
In this method, we define g0(x) to be a generating function if it is monotonic and takes values between 0 and 1 and satisfies the following property,
                    g        0            ⁡              (        x        )              +                  g        0            ⁡              (                  1          x                )              =  1.
Examples of generating functions that resemble a Heaviside function and satisfy the above property are
            g      0        ⁡          (      x      )        =                    2        π            ⁢      a      ⁢                          ⁢              tan        ⁡                  (                      1            x                    )                    ⁢                          ⁢      and      ⁢                          ⁢                        g          0                ⁡                  (          x          )                      =          1              1        +                  x          2                    We seek a series of coefficients an and functions g(x), n=1, . . . , ∞ such that
                              H          ⁡                      (            x            )                          =                                            g              0                        ⁡                          (              x              )                                +                                    ∑                              n                =                1                            ∞                        ⁢                                                  ⁢                                          a                n                            ⁢                                                g                  n                                ⁡                                  (                  x                  )                                                                                        (                  6          ⁢                                          ⁢          ii                )            
For n≧1, we seek functions gn(x) such that the functions satisfy the following properties:
1. gn(x) should have unique inverse Laplace transform in closed-form and should exist for all x.
2. gn(x) should be anti-symmetric in log−x, with
                                          g            n                    ⁡                      (            x            )                          =                  -                                                    g                n                            ⁡                              (                                  1                  x                                )                                      .                                              (                  6          ⁢                                          ⁢          jj                )            3. When x is small, gn(x): xn.4. When x is large, gn(x): x−n.
Properties (1), (3) and (4) are self-explanatory. Property (2) follows from the Heaviside function H(x) and generating function g0(x) satisfying eq. (6hh). At the first iteration, the approximate Heaviside function isH1(x)=g0(x)+a1g1(x)  (6kk)
Therefore,
                                          H            1                    ⁡                      (                          1              x                        )                          =                                            g              0                        ⁡                          (                              1                x                            )                                +                                    a              1                        ⁢                                          g                1                            ⁡                              (                                  1                  x                                )                                                                        (                  6          ⁢                                          ⁢          ll                )            
From our construction, H1(x) and g0(x) satisfy eq. (6hh). Therefore,
                              1          -                                    H              1                        ⁡                          (              x              )                                      =                  1          -                                    g              0                        ⁡                          (              x              )                                +                                    a              1                        ⁢                                          g                1                            ⁡                              (                                  1                  x                                )                                                                        (                  6          ⁢                                          ⁢          mm                )            
Thus, from equations (6kk) and (6mm), we have
            g      1        ⁡          (      x      )        =            -                        g          1                ⁡                  (                      1            x                    )                      ·    A  similar proof follows for any gn(x), n≧1.Case 1:
      Let    ⁢                  ⁢                  g        0            ⁡              (        x        )              =            2      π        ⁢    a    ⁢                  ⁢                  tan        ⁡                  (                      1            x                    )                    .      Its Taylor-series expansion is
                                          2            π                    ⁢          a          ⁢                                          ⁢                      tan            ⁡                          (                              1                x                            )                                      =                                            H              ⁡                              (                x                )                                      -                                                            2                  π                                [                                  x                  -                                                            x                      3                                        3                                    +                                                                                    x                        5                                            5                                        ⁢                                                                                  ⁢                    …                                                  ⁢                                                                  ]                            ⁢                                                          ⁢              for              ⁢                                                          ⁢                                                                                                                                              <          1.                                    (                  6          ⁢                                          ⁢          nn                )            
If we subtract from g0(x) the terms in the Taylor-series proportional to xn(n≧1), written as a function of gn(x), then, we will obtain a function that converges to 1 for |x|<1 and converges to 0 for |x|>1. Since the Taylor-series expansion has only odd-powers of x, we consider
                                          g            n                    ⁡                      (            x            )                          =                                            -                              x                                                      2                    ⁢                                                                                  ⁢                    n                                    -                  1                                                      +                          x                                                2                  ⁢                                                                          ⁢                  n                                +                1                                                                        (                              1                +                                  x                  2                                            )                                      2              ⁢                                                          ⁢              n                                                          (                  6          ⁢                                          ⁢          oo                )            These functions satisfy properties (1)-(4). In addition,
                                          ∑                          k              =              1                        ∞                    ⁢                                          ⁢                                    a              k                        ⁢                                          g                k                            ⁡                              (                x                )                                                    =                              ∑                          k              =              1                        ∞                    ⁢                                          ⁢                                    a              k                        ⁢                                                            -                                      x                                                                  2                        ⁢                                                                                                  ⁢                        k                                            -                      1                                                                      +                                  x                                                            2                      ⁢                                                                                          ⁢                      k                                        +                    1                                                                                                (                                      1                    +                                          x                      2                                                        )                                                  2                  ⁢                                                                          ⁢                  k                                                                                        (                  6          ⁢                                          ⁢          pp                )            Using the series expansion for
  1            (              1        +                  x          2                    )              2      ⁢                          ⁢      k      around x=0, we get
                                          ∑                          k              =              1                        ∞                    ⁢                                          ⁢                                    a              k                        ⁢                                          g                k                            ⁡                              (                x                )                                                    =                              ∑                          k              =              1                        ∞                    ⁢                                          ⁢                                                    a                k                            ⁡                              (                                                      x                                                                  2                        ⁢                                                                                                  ⁢                        k                                            +                      1                                                        -                                      x                                                                  2                        ⁢                        k                                            -                      1                                                                      )                                      ⁡                          [                                                ∑                                      m                    =                    0                                    ∞                                ⁢                                                                  ⁢                                                                            (                                              -                        1                                            )                                        m                                    ⁢                                      (                                                                                                                                                      2                              ⁢                                                                                                                          ⁢                              k                                                        +                            m                            -                            1                                                                                                                                                m                                                                                      )                                    ⁢                                      x                                          2                      ⁢                                                                                          ⁢                      m                                                                                  ]                                                          (                  6          ⁢                                          ⁢          qq                )            Matching the coefficients for x2n−1 in eqns. (43) and (46) yields
                              a          n                =                                                            (                                  -                  1                                )                                            n                -                1                                                                    2                ⁢                                                                  ⁢                n                            -              1                                +                                    ∑                              k                =                1                                            n                -                1                                      ⁢                                                  ⁢                                                            (                                      -                    1                                    )                                                  n                  -                  k                                            ⁢                                                                    a                    k                                    ⁡                                      [                                                                  (                                                                                                                                            n                                +                                k                                -                                2                                                                                                                                                                                                        n                                -                                k                                -                                1                                                                                                                                    )                                            +                                              (                                                                                                                                            n                                +                                k                                -                                1                                                                                                                                                                                                        n                                -                                k                                                                                                                                    )                                                              ]                                                  .                                                                        (                  6          ⁢                                          ⁢          rr                )            
The first three coefficients are
            a      1        =          2      π        ,            a      2        =                            8          3                ⁢                  a          1                ⁢                                  ⁢        and        ⁢                                  ⁢                  a          3                    =                                    128            15                    ⁢                                    a              1                        .                                                  ⁢            Let                    ⁢                                          ⁢          τ                =                              t                          T              c                                .                    The inverse Laplace transforms of the first three terms in the series expansion are
                    ⁢                            L                      -            1                          ⁡                  [                                    x              -                              x                3                                                                    (                                  1                  +                                      x                    2                                                  )                            2                                ]                    =                        1                      T            c                          ⁡                  [                                    τ              ⁢                                                          ⁢                              sin                ⁡                                  (                  τ                  )                                                      -                          cos              ⁡                              (                τ                )                                              ]                                        ⁢                            L                      -            1                          ⁡                  [                                                    x                3                            -                              x                5                                                                    (                                  1                  +                                      x                    2                                                  )                            4                                ]                    =                        1                      T            c                          ⁡                  [                                    1              24                        ⁢                          (                                                                    -                    6                                    ⁢                                      τ                    2                                    ⁢                                      cos                    ⁡                                          (                      τ                      )                                                                      -                                  6                  ⁢                  τ                  ⁢                                                                          ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                      +                                                      τ                    3                                    ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                                  )                                ]                                        L                  -          1                    ⁡              [                                            x              5                        -                          x              7                                                          (                              1                +                                  x                  2                                            )                        6                          ]              =                            1                      T            c                          ⁡                  [                                    1              1920                        ⁢                          (                                                30                  ⁢                                                                          ⁢                                      τ                    2                                    ⁢                                      cos                    ⁡                                          (                      τ                      )                                                                      -                                  15                  ⁢                                      τ                    4                                    ⁢                                      cos                    ⁡                                          (                      τ                      )                                                                      -                                  30                  ⁢                  τ                  ⁢                                                                          ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                      -                                  55                  ⁢                                      τ                    3                                    ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                      +                                                      τ                    5                                    ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                                  )                                ]                    .      
A general expression for the inverse Laplace transform for gn(x) can be obtained from the following:
                                          L                          -              1                                ⁡                      [                                                            x                  r                                -                                  x                                      r                    +                    2                                                                                                (                                      1                    +                                          x                      2                                                        )                                                  r                  +                  1                                                      ]                          =                                            τ                              r                -                1                                                                    T                c                            ⁢                              Γ                ⁡                                  (                  r                  )                                            ⁢                              Γ                ⁡                                  (                                      r                    +                    2                                    )                                                              ⁡                      [                                                            τ                  2                                ⁢                                                      Γ                    ⁡                                          (                      r                      )                                                        1                                ⁢                                                      F                    2                                    ⁡                                      (                                                                                            r                          +                          1                                                ;                                                                              r                            2                                                    +                          1                                                                    ,                                                                                                    r                            2                                                    +                                                      3                            2                                                                          ;                                                  -                                                                                    τ                              2                                                        4                                                                                                                )                                                              -                                                                    Γ                    ⁡                                          (                                              r                        +                        2                                            )                                                        1                                ⁢                                                      F                    2                                    ⁡                                      (                                                                                            r                          +                          1                                                ;                                                                              r                            2                                                    +                                                      1                            2                                                                                              ,                                                                        r                          2                                                ;                                                  -                                                                                    τ                              2                                                        4                                                                                                                )                                                                        ]                                              (                  6          ⁢                                          ⁢          ss                )            where 1F2 refers to the generalized hypergeometric function.Case 2:
      Let    ⁢                  ⁢                  g        0            ⁡              (        x        )              =            1              1        +                  x          2                      .  Its Taylor-series expansion (around x=0) is
                              1                      1            +                          x              2                                      =                              H            ⁡                          (              x              )                                -                      x            2                    +                      x            4                    -                                    x              6                        ⁢                                                  ⁢            …                                              (                  6          ⁢                                          ⁢          tt                )            Since the series expansion has only even powers of x, we consider gn(x) of the form,
                                                        g              n                        ⁡                          (              x              )                                =                                                    x                                                      2                    ⁢                                                                                  ⁢                    n                                    -                  2                                            -                              x                                  2                  ⁢                                                                          ⁢                  n                                                                                    (                                  1                  +                                      x                    2                                                  )                                                              2                  ⁢                                                                          ⁢                  n                                -                1                                                    ,                  n          =          1                ,        2        ,        …                            (                  6          ⁢                                          ⁢          uu                )            These functions satisfy properties (1)-(4). Using series expansion of
  1            (              1        +                  x          2                    )                      2        ⁢                                  ⁢        n            -      1      around x=0, we get,
                                              ⁢                                            ∑                              k                =                1                            ∞                        ⁢                                                  ⁢                                          a                k                            ⁢                                                g                  k                                ⁡                                  (                  x                  )                                                              =                                                    ∑                                  k                  =                  1                                ∞                            ⁢                                                          ⁢                                                a                  k                                ⁢                                                                            -                                              x                                                                              2                            ⁢                                                                                                                  ⁢                            k                                                    -                          2                                                                                      +                                          x                                              2                        ⁢                                                                                                  ⁢                        k                                                                                                                        (                                              1                        +                                                  x                          2                                                                    )                                                                                      2                        ⁢                                                                                                  ⁢                        k                                            -                      1                                                                                            =                                              (                  6          ⁢                                          ⁢          vv                )                                          ∑                      k            =            1                    ∞                ⁢                                  ⁢                                            a              k                        ⁡                          (                                                x                                                            2                      ⁢                                                                                          ⁢                      k                                        -                    2                                                  -                                  x                                      2                    ⁢                                                                                  ⁢                    k                                                              )                                ⁡                      [                                          ∑                                  m                  =                  0                                ∞                            ⁢                                                          ⁢                                                                    (                                          -                      1                                        )                                    m                                ⁢                                  (                                                                                                                                          2                            ⁢                                                                                                                  ⁢                            k                                                    +                          m                          -                          2                                                                                                                                    m                                                                              )                                ⁢                                  x                                      2                    ⁢                                                                                  ⁢                    m                                                                        ]                                              (                  6          ⁢          ww                )            Matching the coefficients for x2n−2 in eqns. (49) and (52) yields
                              a          n                =                                            (                              -                1                            )                                      n              -              1                                -                                    ∑                              k                =                1                                            n                -                1                                      ⁢                                                  ⁢                                                            (                                      -                    1                                    )                                                  n                  -                  k                                            ⁢                                                                    a                    k                                    ⁡                                      [                                                                  (                                                                                                                                            n                                +                                k                                -                                2                                                                                                                                                                                                        n                                -                                k                                                                                                                                    )                                            +                                              (                                                                                                                                            n                                +                                k                                -                                3                                                                                                                                                                                                        n                                -                                k                                -                                1                                                                                                                                    )                                                              ]                                                  .                                                                        (                  6          ⁢                                          ⁢          xx                )            
The first three coefficients are a1=0, a2=−1 and a3=−3. Let
  τ  =            t              T        c              .  The inverse Laplace transforms of the first three terms in the series expansion are
                    ⁢                            L                      -            1                          ⁡                  [                                    1              -                              x                2                                                    (                              1                +                                  x                  2                                            )                                ]                    =                        1                      T            c                          ⁡                  [                                    2              ⁢                                                          ⁢                              sin                ⁡                                  (                  τ                  )                                                      -                          δ              ⁡                              (                τ                )                                              ]                                        ⁢                            L                      -            1                          ⁡                  [                                                    x                2                            -                              x                4                                                                    (                                  1                  +                                      x                    2                                                  )                            3                                ]                    =                        1                      T            c                          ⁡                  [                                    1              4                        ⁢                          (                                                                    τ                    2                                    ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                      -                                  sin                  ⁡                                      (                    τ                    )                                                  -                                  3                  ⁢                  τ                  ⁢                                                                          ⁢                                      cos                    ⁡                                          (                      τ                      )                                                                                  )                                ]                                        L                  -          1                    ⁡              [                                            x              4                        -                          x              6                                                          (                              1                +                                  x                  2                                            )                        5                          ]              =                            1                      T            c                          ⁡                  [                                    1              192                        ⁢                          (                                                                    τ                    4                                    ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                      -                                  10                  ⁢                                      τ                    3                                    ⁢                                      cos                    ⁡                                          (                      τ                      )                                                                      -                                  21                  ⁢                                      τ                    2                                    ⁢                                      sin                    ⁡                                          (                      t                      )                                                                      -                                  3                  ⁢                                                                          ⁢                                      sin                    ⁡                                          (                      τ                      )                                                                      +                                  3                  ⁢                  τ                  ⁢                                                                          ⁢                                      cos                    ⁡                                          (                      τ                      )                                                                                  )                                ]                    .      
It can be shown that the Taylor-series expansion of the generating function in terms of anti-symmetric higher-order polynomials systematically leads to convergence of the generating function to a Heaviside function in log(x) space.
As explained in previously incorporated U.S. Ser. No. 13/333,232, the above describes methods for computing linear functionals of the distribution function without first computing the distribution of relaxation times. These methods involve a linear transform of the measured data using integral transforms. Different linear functionals of the distribution function can be obtained by choosing appropriate functions in the integral transforms. This approach can be used such that integral transforms can be computed on data corresponding to longitudinal relaxation time (T1), and such that integral transforms can be computed on data corresponding to diffusion coefficient (D). In addition, this approach can be extended to multiple dimensions.
Using the techniques given in previously incorporated U.S. Ser. No. 13/333,232, it is possible to estimate the uncertainty in the parameters. Let the discretized version of the linear transform be denoted by A=kTG, where k is the discretization of the k(t) in equation (4) and G refers to the discretization of the measured data indications. Let σε2 be the variance of the vector of measurement indications G. Then the variance of A is given byσA2=σε2∥k∥2  (7)